a; \(x-33=28\)

\(x=28+33\)

\(x=61\)

Vậy \(x=61\)

b; \(x+55\) = 122

\(x=122-55\)

\(x=\) 67

Vậy \(x=67\)

c; \(x\times34\) = 37

\(x\) = 37 : 34

\(x=\frac{37}{34}\)

Vậy \(x=\frac{37}{34}\)

d; \(x:23\) = 7

\(x=7\times23\)

\(x=161\)

Vậy \(x=161\)

e; \(x^2=81\)

\(x=9^2\)

\(\left[\begin{array}{l}x=-9\\ x=9\end{array}\right.\)

Vậy \(x\in\) {-9; 9}

f; (\(x-3)^3\) = 27

(\(x-3)^3=3^3\)

\(x-3=3\)

\(x=3+3\)

\(x=6\)

Vậy \(x=6\)

a; \(x - 33 = 28\)

\(x = 28 + 33\)

\(x = 61\)

Vậy \(x = 61\)

b; \(x + 55\) = 122

\(x = 122 - 55\)

\(x =\) 67

Vậy \(x = 67\)

c; \(x \times 34\) = 37

\(x\) = 37 : 34

\(x = \frac{37}{34}\)

Vậy \(x = \frac{37}{34}\)

d; \(x : 23\) = 7

\(x = 7 \times 23\)

\(x = 161\)

Vậy \(x = 161\)

e; \(x^{2} = 81\)

\(x = 9^{2}\)

\(\left[\right. x = - 9 \\ x = 9\)

Vậy \(x \in\) {-9; 9}

f; (\(x - 3 \left.\right)^{3}\) = 27

(\(x - 3 \left.\right)^{3} = 3^{3}\)

\(x - 3 = 3\)

\(x = 3 + 3\)

\(x = 6\)

Vậy \(x = 6\)

a: \(\left\lbrack\left(4^{60}:4^{58}+3^2\right):25-1^{100}\right\rbrack\cdot5^{500}\)

\(=\left\lbrack\frac{\left(4^2+3^2\right)}{25}-1\right\rbrack\cdot5^{500}\)

\(=\left(\frac{25}{25}-1\right)\cdot5^{500}=0\)

b: \(\left\lbrace\left\lbrack\left(10^2-6^2\right):8+1^{100}\right\rbrack:3^2+2^4\right\rbrace:17\)

\(=\frac{\left\lbrace\left\lbrack\frac{\left(100-36\right)}{8}+1\right\rbrack:9+16\right\rbrace}{17}\)

\(=\frac{\left\lbrack64:8+1\right\rbrack:9+16}{17}=\frac{\left(8+1\right):9+16}{17}=\frac{1+16}{17}=1\)

Bài 1:

a: 2x+3=15

=>2x=15-3=12

=>\(x=\frac{12}{2}=6\)

b: 28-3x=13

=>\(3x=28-13=15\)

=>\(x=\frac{15}{3}=5\)

c: \(\left(x-1954\right)\cdot5=50\)

=>x-1954=50:5=10

=>x=10+1954=1964

d: 30(60-x)=30

=>60-x=1

=>x=60-1=59

Bài 2:

a: x+99:3=55

=>x+33=55

=>x=55-33=22

b: (x-25):15=20

=>\(x-25=20\cdot15=300\)

=>x=300+25=325

c: \(7\left(3x-15\right)=42\)

=>\(3x-15=\frac{42}{7}=6\)

=>3x=15+6=21

=>\(x=\frac{21}{3}=7\)

d: \(x\left(x+1\right)=2+4+6+\cdots+2500\)

=>\(x\left(x+1\right)=2\left(1+2+3+\cdots+1250\right)\)

=>\(x\left(x+1\right)=2\cdot1250\cdot\frac{1251}{2}=1250\cdot1251\)

=>\(x^2+x-1250\cdot1251=0\)

=>(x+1251)(x-1250)=0

=>\(\left[\begin{array}{l}x+1251=0\\ x-1250=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-1251\\ x=1250\end{array}\right.\)

Bài 1:

a: \(46\cdot99=46\cdot\left(100-1\right)=46\cdot100-46\)

=4600-46

=4554

b: \(47\cdot98=47\cdot\left(100-2\right)=47\cdot100-47\cdot2\)

=4700-94

=4604

c: \(18\cdot19=18\left(20-1\right)=18\cdot20-18\)

=360-18

=342

d: \(24\cdot198=24\left(200-2\right)\)

\(=24\cdot200-24\cdot2\)

=4800-48

=4752

Bài 2:

a: \(1800\cdot5=18\cdot100\cdot5=90\cdot100=9000\)

b: \(36\cdot25=25\cdot4\cdot9=100\cdot9=900\)

c: 36600:50=3660:5=732

d: 220000:5000=220:5=44

sạc xong rồi học đk

c: \(\left(x-1\right)^3=\left(-9\right)^3\)

=>x-1=-9

=>x=-9+1=-8

f: \(3x-2^3=7+\left(-9\right)\)

=>3x-8=7-9=-2

=>3x=-2+8=6

=>x=2

Gọi tổng chi phí cho chuyến đi là x(đồng)

(Điều kiện: x>0)

Số tiền mỗi bạn phải đóng ban đầu là \(\frac{x}{40}\) (đồng)

Số tiền mỗi bạn phải đóng thực tế là \(\frac{x}{36}\) (đồng)

Mỗi bạn tham gia còn lại phải đóng thêm 25000 đồng nên ta có:

\(\frac{x}{36}-\frac{x}{40}=25000\)

=>\(\frac{x}{360}=25000\)

=>\(x=25000\cdot360=9000000\) (đồng)

vậy: Tổng chi phí cho chuyến đi là 9000000(đồng)

Bài 3:

4; 45 + 5\(x\) = 10\(^3\): 10

45 + 5\(x\) = 100

5\(x\) = 100 - 45

5\(x\) = 55

\(x\) = 55 : 5

\(x\) = 11

Vậy \(x=11\)

5; 4\(x\) - 20 = 2\(^5\) : 2\(^2\)

4\(x\) - 20 = 2\(^3\)

4\(x\) = 8 + 20

4\(x\) = 28

\(x\) = 28 : 4

\(x=7\)

Vậy \(x=7\)

Bài 4:

1; 82 - (25 + 4\(x^{}\)) = 17

25 + 4\(x\) \(^{}\) = 82 - 17

4\(x^{}\) = 65 - 25

4\(x^{}\) = 40

\(x=40:4\)

\(x\) = 10

Vậy \(x=10\)

2; 71 - (24 + 3\(x\)) = 24

24 + 3\(x\) = 71 - 24

24 + 3\(x\) = 47

3\(x\) = 47 - 24

3\(x\) = 23

\(x\) = 23 : 3

Vậy \(x=\frac{23}{3}\)

3; 145 - (125 + \(x\)) = 12

125 + \(x\) = 145 - 12

125 + \(x\) = 133

\(x\) = 133 - 125

\(x\) = 8

Vậy \(x=8\)